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%%文档的题目、作者与日期
%\author{王立庆（2020级数学与应用数学1班） }
\author{学号 \underline{\hspace{4cm}} \hspace{1cm} 姓名 \underline{\hspace{4cm}} }
\title{常微分方程期中考试解答}
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\date{2023 年 11 月 23 日}
%\date{March 9, 2021}

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\begin{document}

\maketitle

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\begin{enumerate}

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\item  %第1题
判断下列方程是否为恰当方程，并对恰当方程求解：$$(y^2+2)\cos(x)dx + 2y[\sin(x)+\cos(y)]dy=0. $$

\vspace{0.2cm}

{\color{red}解答：
%\begin{enumerate}[label={(\arabic*)}]
\begin{enumerate}[label={\arabic*.}]
\item  原方程是 $Pdx+Qdy=0$, 其中 $P=(y^2+2)\cos(x)$, $Q=2y[\sin(x)+\cos(y)]$. 
\item  计算偏导数可得 $P'_y=2y\cos(x)$ 与 $Q'_x=2y\cos(x).$
\item  因为 $P'_y=Q'_x$, 所以这是恰当方程。
\item  设 $d\Phi = Pdx+Qdy$, 则有 $\Phi'_x=P$ 与 $\Phi'_y=Q$. 
\item  由 $\Phi'_x=P=(y^2+2)\cos(x)$ 可得 $\Phi = (y^2+2)\sin(x) + \varphi(y)$, 其中 $\varphi(y)$ 待定。
\item  代入 $\Phi'_y=Q=2y[\sin(x)+\cos(y)]$ 可得 $\varphi'(y)=2y\cos(y)$.  
\item  积分可得 $\varphi(y)=2y\sin(y)+2\cos(y)+C$. 
\item  原方程的通解为 $(y^2+2)\sin(x) + 2y\sin(y) + 2\cos(y) + C=0$, 其中 $C$ 是任意常数。
\end{enumerate} 

}

\vspace{2cm}

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\item  %第2题
求解下述微分方程的初值问题：$$\sqrt{1+x^2}\frac{dy}{dx}=xy^3,\,\, y(0)=2. $$

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}[label={\arabic*.}]
\item  分离变量可得 $\frac{dy}{y^3} = \frac{xdx}{\sqrt{1+x^2}}$, 此即 $2y^{-3}dy = (1+x^2)^{-1/2}d(1+x^2)$. 
\item  两边积分可得 $-y^{-2} = 2(1+x^2)^{1/2}+C$. 此即 $y^2=\frac{-1}{2\sqrt{1+x^2}+C}$. 
\item  代入初值条件 $x_0=0, y_0=2$ 可得 $4=\frac{-1}{2+C}$. 求得 $8+4C=-1$, $C=-\frac{-9}{4}$. 
\item  因此初值问题的解为 $y^2=\frac{4}{9-8\sqrt{1+x^2}}, y>0$.
\end{enumerate} 

}


\vspace{1cm}

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\item  %第3题
使用变量代换 $u=\sin(y)$, 求解微分方程：$$\frac{dy}{dx}=\frac{\cos(x)}{\cos(y)} + \tan(y). $$

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}[label={\arabic*.}]
\item  方程两边同乘以 $\cos(y)$ 可得 $\cos(y)\frac{dy}{dx} = \cos(x) + \sin(y)$. 
\item  使用变量代换 $u=\sin(y)$, 方程化为 $\frac{du}{dx}=\cos(x)+u$.  
\item  在方程 $\frac{du}{dx} - u = \cos(x)$ 两边同乘以积分因子 $m(x)=e^{-x}$ 可得 
$e^{-x}[\frac{du}{dx} - u] = e^{-x}\cos(x)$. 
\item  两边积分可得 $e^{-x}u=\int e^{-x}\cos(x)dx = \frac{1}{2}e^{-x}\sin(x) -\frac{1}{2} e^{-x}\cos(x) +C$. 
\item  因此原方程的解为 $u=\frac{1}{2}\sin(x) -\frac{1}{2} \cos(x) +Ce^x$, 其中 $C$ 是任意常数。
\item  代回原变量可得 $\sin(y)=\frac{1}{2}\sin(x) -\frac{1}{2} \cos(x) +Ce^x$. 
\end{enumerate} 


}

\vspace{2cm}

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\item  %第4题
考虑变量代换 $u=\frac{y'}{y}$, 将微分方程 $y''+ay'+by=0$ 化成关于 $u(x)$ 的微分方程。%，并求解该方程。

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}[label={\arabic*.}]
\item  由 $u=\frac{y'}{y}$ 可得 $y'=uy$, 对 $x$ 求导可得 $y''=u'y+uy'$. 
\item  将上式的 $y'$ 由 $y'=uy$ 代替，可得 $y''=u'y+u^2y$. 
\item  原方程化为 $(u'y+u^2y)+auy+by=0$. 
\item  约去 $y$ 可得 $u'+u^2+au+b=0$. 
\item  分离变量可得 $\frac{du}{u^2+au+b} = -dx$. 
\item  两边积分可得 $\int \frac{du}{u^2+au+b} = \int -dx$. 
\end{enumerate} 


}

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\item  %第5题
考虑初值问题 $\frac{dy}{dx} = F(x,y),\, y(0)=0$, 其中函数 $F(x,y)$ 在矩形区域 $[0,1]\times (-\infty,\infty)$ 分区域定义。 
\begin{eqnarray*}
F(x,y) = \left\{\begin{array}{ll}
0, & x=0, -\infty<y<\infty, \\
2x, & 0<x\le 1, -\infty<y<0, \\
2x-\frac{4y}{x}, & 0<x\le 1, 0\le y<x^2, \\
-2x, & 0<x\le 1, x^2\le y<\infty.
\end{array}\right.
\end{eqnarray*}
计算该初值问题的皮卡序列的前四个函数。

\vspace{0.2cm}

{\color{red}解答：根据皮卡序列的定义，可得
\begin{eqnarray*}
y_0(x) &=& y(0)=0, \\ 
y_1(x) &=& y_0 + \int_0^x F(t,y_0(t))dt = \int_0^x F(t,0)dt = \int_0^x (2t)dt = x^2, \\  
y_2(x) &=& y_0 + \int_0^x F(t,y_1(t))dt = \int_0^x F(t,t^2)dt = \int_0^x (-2t)dt = -x^2, \\  
y_3(x) &=& y_0 + \int_0^x F(t,y_2(t))dt = \int_0^x F(t,-t^2)dt = \int_0^x (2t)dt = x^2. 
\end{eqnarray*}

}

\vspace{2cm}

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\item  %第6题
记 $p=\frac{dy}{dx}$, 求微分方程 $xp^3-yp^2-1=0$ 的所有解。

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}[label={\arabic*.}]
\item  方程两边对 $x$ 求导可得 $p^3+3xp^2p'-pp^2-2ypp'=0$. 化简可得 $3xp^2p'-2ypp'=0$. 
\item  因式分解可得 $(3xp-2y)pp'=0$. 
\item  由 $3xp=2y$ 可得 $p=\frac{2y}{3x}$, 代入原方程可得 $4y^3+27x^2=0$. \item  将 $p=0$ 代入原方程，可知原方程不成立。
\item  由 $p'=0$ 可得 $p=c$ 是任意常数，代入原方程可得 $xc^3-yc^2-1=0$. 这是一族直线。
\item  原方程的通解为 $y=cx-\frac{1}{c^2}$. 
%\item  为验证特解 $4y^3+27x^2=0$ 是奇解，对其进行隐函数求导，可得 $12y^2y'+54x=0$. 
%\item  因此特解 $4y^3+27x^2=0$ 上的每点的切线斜率是 $y'=\frac{-54x}{12y^2}=\frac{-9x}{2y^2}$. 
%\item  联立特解与通解，可得 
\item  原方程的特解为 $y=\sqrt[3]{-27x^2/4} = -\frac{3}{2}\sqrt[3]{2x^2}$. 

\end{enumerate} 


}

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\item  %第7题
\begin{enumerate}
\item  将微分方程 $\frac{d^2x}{dt^2} = -\sin(x)$ 写成一阶微分方程组的形式。
\item  设 $v=\frac{dx}{dt}$, 将微分方程 $\frac{d^2x}{dt^2} = -\sin(x)$ 化为关于 $v$ 与 $x$ 的微分方程，并求解该微分方程。
\end{enumerate}

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}
\item  记 $y=\frac{dx}{dt}$, 则原方程可写为 
$%\begin{eqnarray*}
\left\{\begin{array}{rcl}
\frac{dx}{dt} &=& y, \\ 
\frac{dy}{dt} &=& -\sin(x). 
\end{array}\right.
$. %\end{eqnarray*}

\item  
\begin{enumerate}[label={\arabic*.}]
\item  在 $v=\frac{dx}{dt}$ 的两边对 $t$ 求导，可得 $\frac{dv}{dt} = \frac{d^2x}{dt^2}$. 
\item  根据链式法则可得 $\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}.$
\item  因此原方程可以化为 $v\frac{dv}{dx}=-\sin(x)$. 
\item  分离变量可得 $vdv = -\sin(x)dx$. 
\item  积分可得 $v^2 = 2\cos(x)+C$, 其中 $C$ 是任意常数。
\end{enumerate} 

\end{enumerate} 

}

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\item  %第8题
求经过点 $(1,1)$ 的曲线，使其与双曲线族 $x^2-4y^2=C$ 都垂直。
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{\color{red}解答：
\begin{enumerate}[label={\arabic*.}]
\item  对 $x^2-4y^2=C$ 微分可得 $2xdx - 8ydy=0$. 
\item  写成微分方程可得 $\frac{dy}{dx}=\frac{x}{4y}$. 
\item  正交轨线族的微分方程为 $\frac{dy}{dx}=\frac{-4y}{x}$. 
\item  分离变量可得 $\frac{dy}{y}=\frac{-4dx}{x}$. 
\item  积分可得 $\ln |y| = -4\ln |x| + C$. 
\item  上述正交轨线族中，经过点 $(1,1)$ 的曲线方程为 $\ln y = -4\ln x$, 即 $y=\frac{1}{x^4}$. 
\end{enumerate} 

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\includegraphics [height=5cm, width=8cm]{ode-midterm-3-problem-8.png}
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}

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\end{enumerate}


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\end{document}

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